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A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s?
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?A particle of mass 50 g is projected horizontallyfrom the top of a to...
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?A particle of mass 50 g is projected horizontallyfrom the top of a to...
Given:
- Mass of the particle (m) = 50 g = 0.05 kg
- Height of the tower (h) = 50 m
- Initial velocity (u) = 20 m/s
- Acceleration due to gravity (g) = 10 m/s²
(1) Velocity at the instant t = 2 s:
- We can use the formula of motion to calculate the final velocity (v) at a given time (t).
- The formula is v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
- Substituting the given values, we have v = 20 + 10 * 2 = 40 m/s.
Therefore, the velocity at t = 2 s is 40 m/s.
(2) Position at the instant t = 3 s:
- To find the position at a given time, we can use the formula of motion: s = ut + (1/2)gt².
- Here, s is the position, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
- Substituting the given values, we have s = 20 * 3 + (1/2) * 10 * (3)² = 60 + 45 = 105 m.
Therefore, the position at t = 3 s is 105 m.
(3) Velocity just before hitting the ground:
- The time taken by the particle to reach the ground can be found using the formula h = ut + (1/2)gt², where h is the height of the tower.
- Substituting the given values, we have 50 = 20t + (1/2) * 10 * t².
- This equation can be simplified to t² + 4t - 10 = 0 by rearranging terms.
- Solving this quadratic equation, we get t ≈ 1.16 s (ignoring the negative value).
- The final velocity just before hitting the ground is v = u + gt, where u is the initial velocity, g is the acceleration due to gravity, and t is the time.
- Substituting the given values, we have v = 20 + 10 * 1.16 ≈ 31.6 m/s.
Therefore, the velocity just before hitting the ground is approximately 31.6 m/s.
(4) Change in velocity in 2 s:
- The change in velocity can be calculated using the formula Δv = v - u, where Δv is the change in velocity, v is the final velocity, and u is the initial velocity.
- Substituting the given values, we have Δv = 40 - 20 = 20 m/s.
Therefore, the change in velocity in 2 seconds is 20 m/s.
In summary:
(1) Velocity at t = 2 s: 40 m/s
(2) Position at t = 3 s: 105 m
(3) Velocity just before hitting the ground: approximately 31.6 m/s
(4) Change in velocity in 2 s: 20 m/s
- Mass of the particle (m) = 50 g = 0.05 kg
- Height of the tower (h) = 50 m
- Initial velocity (u) = 20 m/s
- Acceleration due to gravity (g) = 10 m/s²
(1) Velocity at the instant t = 2 s:
- We can use the formula of motion to calculate the final velocity (v) at a given time (t).
- The formula is v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
- Substituting the given values, we have v = 20 + 10 * 2 = 40 m/s.
Therefore, the velocity at t = 2 s is 40 m/s.
(2) Position at the instant t = 3 s:
- To find the position at a given time, we can use the formula of motion: s = ut + (1/2)gt².
- Here, s is the position, u is the initial velocity, g is the acceleration due to gravity, and t is the time.
- Substituting the given values, we have s = 20 * 3 + (1/2) * 10 * (3)² = 60 + 45 = 105 m.
Therefore, the position at t = 3 s is 105 m.
(3) Velocity just before hitting the ground:
- The time taken by the particle to reach the ground can be found using the formula h = ut + (1/2)gt², where h is the height of the tower.
- Substituting the given values, we have 50 = 20t + (1/2) * 10 * t².
- This equation can be simplified to t² + 4t - 10 = 0 by rearranging terms.
- Solving this quadratic equation, we get t ≈ 1.16 s (ignoring the negative value).
- The final velocity just before hitting the ground is v = u + gt, where u is the initial velocity, g is the acceleration due to gravity, and t is the time.
- Substituting the given values, we have v = 20 + 10 * 1.16 ≈ 31.6 m/s.
Therefore, the velocity just before hitting the ground is approximately 31.6 m/s.
(4) Change in velocity in 2 s:
- The change in velocity can be calculated using the formula Δv = v - u, where Δv is the change in velocity, v is the final velocity, and u is the initial velocity.
- Substituting the given values, we have Δv = 40 - 20 = 20 m/s.
Therefore, the change in velocity in 2 seconds is 20 m/s.
In summary:
(1) Velocity at t = 2 s: 40 m/s
(2) Position at t = 3 s: 105 m
(3) Velocity just before hitting the ground: approximately 31.6 m/s
(4) Change in velocity in 2 s: 20 m/s
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?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s?
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?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about ?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s?.
?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about ?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?A particle of mass 50 g is projected horizontallyfrom the top of a tower of height 50 m with avelocity 20 m/s. If g = 10 m/s, then find the :-(1) Velocity at the instant t = 2 s(2) Position at the instant t = 3 s(3) Velocity just before hitting the ground(4) Change in velocity in 2 s?.
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